【CodeForces】446C Number of Ways

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传送门：【CodeForces】446C Number of Ways

题目分析：首先求出所有相加为2/3sum的位置，用后缀和求出从这个点开始到结尾内相加为2/3sum的点的个数suffix_sum[ i ]。最后再for一遍，每到一个1/3sum的点i，ans += suffx_sum[ i + 1]。

最后输出ans即可。

代码如下：

#include <cmath>#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;typedef long long LL ;#define REP( i , a , b ) for ( int i = a ; i < b ; ++ i )#define REV( i , a , b ) for ( int i = a ; i >= b ; -- i )#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define CLR( a , x ) memset ( a , x , sizeof a )#define CPY( a , x ) memcpy ( a , x , sizeof a )const int MAXN = 500005 ;bool vis[MAXN] ;int suffix_sum[MAXN] ;int a[MAXN] ;int n ;void solve () {LL sum = 0 ;FOR ( i , 1 , n ) {scanf ( "%d" , &a[i] ) ;sum += a[i] ;}if ( sum % 3 ) printf ( "0\n" ) ;else {CLR ( vis , 0 ) ;CLR ( suffix_sum , 0 ) ;LL tmp = 0 , ans = 0 ;FOR ( i , 1 , n ) {tmp += a[i] ;if ( tmp * 3 == sum * 2 ) vis[i] = 1 ;}REV ( i , n - 1 , 1 ) suffix_sum[i] = suffix_sum[i + 1] + vis[i] ;tmp = 0 ;FOR ( i , 1 , n - 1 ) {tmp += a[i] ;if ( tmp * 3 == sum ) ans += suffix_sum[i + 1] ;}printf ( "%I64d\n" , ans ) ;}}int main () {while ( ~scanf ( "%d" , &n ) ) solve () ;return 0 ;}

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